正态分布乘积等于正态分布的证明_正态分布的乘积

正态分布乘积

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Tina Memo No. 2003-003

Internal Report

Products and Convolutions of Gaussian Probability Density

Functions

P.A. Bromiley

Last updated

14 / 8 / 2014

Imaging Sciences Research Group, Institute of Population Health,

School of Medicine, University of Manchester,

Stopford Building, Oxford Road,

Manchester, M13 9PT.

Products and Convolutions of Gaussian Probability

Density Functions

P. A. Bromiley

Imaging Sciences Research Group, Institute of Population Health,

School of Medicine, University of Manchester,

Manchester, M13 9PT, UK

paul.bromiley@manchester.ac.uk

Abstract

It is well known that the product and the convolution of Gaussian probability density functions (PDFs)

are also Gaussian functions. This document provides proofs of this for several cases; the product of

two univariate Gaussian PDFs, the product of an arbitrary number of univariate Gaussian PDFs, the

product of an arbitrary number of multivariate Gaussian PDFs , and the convolution of two univari-

ate Gaussian PDFs. These results are useful in calculating the eﬀects of smoothing applied as an

intermediate step in various algorithms.

1 The Product of Two Univariate Gaussian PDFs

Let f(x) and g(x) be Gaussian PDFs with arbitrary means µ

and µ

and standard deviations σ

and σ

f(x) =

√

2πσ

−

(x−µ

)

2σ

and g(x) =

√

2πσ

−

(x−µ

)

2σ

Their product is

f(x)g(x) =

2πσ

−



(x−µ

)

2σ

(x−µ

)

2σ



Examine the term in the exponent

β =

(x − µ

)

2σ

(x − µ

)

2σ

Expanding the two quadratics and collecting te r ms in powers of x gives

β =

(σ

+ σ

− 2(µ

+ µ

)x + µ

+ µ

2σ

Dividing through by the coeﬃcient of x

gives

β =

− 2

+µ

+σ

x +

+µ

+σ

This is again a quadratic in x, and so Eq. 2 is a Gaussian function. Compare the terms in Eq. 5 to a the usual

Gaussian form

P (x) =

√

2πσ

−

(x−µ)

2σ

√

2πσ

−

−2µx+µ

)

2σ

Since a term ǫ that is independent of x can be added to complete th e square in β, this is suﬃc ent to complete the

proof in cases where the normalisation can be ignored. The product of two Gaussian PDFs is proportional to a

Gaussian PDF with a mean that is half the coeﬃcient of x in Eq. 5 and a standard deviation that is the square

root of half of the denominator i.e.

+ σ

and µ

+ µ

+ σ

i.e. the variance σ

is twice the harmonic mean of the individual variances σ

and σ

, and the mean µ

is the

sum of the individual means µ

and µ

weighted by their variances. In general, the product is not itself a PDF

as, due to the presence of the scaling factor, it will not have the corr e ct nor malis ation.

The product f(x)g(x) can now be written in the usual Gaus s ian form directly, with an unknown scaling constant

(this may be suﬃcient in cases where renormalisation can be applied). Alternatively, proceeding from Eq. 5,

suppose that ǫ is the term required to complete the square in β i.e.

ǫ =



+µ

+σ



−



+µ

+σ



2σ

(σ

+σ

)

= 0

Adding this term to β gives

β =

− 2x

+µ

+σ



+µ

+σ



2σ

(σ

+σ

)

+µ

+σ

−



+µ

+σ



2σ

(σ

+σ

)

After some manipulation, this reduces to

β =



x −

+µ

+σ



+σ

(µ

− µ

)

2(σ

+ σ

)

(x − µ

)

2σ

(µ

− µ

)

2(σ

+ σ

)

Substituting back into Eq. 2 gives

f(x)g(x) =

2πσ

exp

−

(x − µ

)

2σ

exp

−

(µ

− µ

)

2(σ

+ σ

)

Multiplying by σ

/σ

and rearranging gives

√

2πσ

exp

−

(x − µ

)

2σ

2π(σ

+ σ

)

exp

−

(µ

− µ

)

2(σ

+ σ

)

Therefore, the product of two Gaussians PDFs f(x) and g(x) is a scaled Gaussian PDF

f(x)g(x) =

√

2πσ

exp

−

(x − µ

)

2σ

where

+ σ

and µ

+ µ

+ σ

(1)

and the scaling factor S is itself a Gaussian PDF on both µ

and µ

with standard deviation

+ σ

2π(σ

+ σ

)

exp

−

(µ

− µ

)

2(σ

+ σ

)

These can be wr i tt en more conveniently as

, µ

and S

2π

f g

exp

−

(µ

− µ

)

(2)

It is much easier to generate a proof by induction for the scaling factor of products of larger numbers of Gaussians

if it is written in the form of a sum of terms, each of which involves a single subscript i.e. the parameters of a

single Gaussian PDF. Appendix A provides the necessary proof, giving

2π

f g

exp

−

(3)

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